Question

The astronomical unit (AU, equal to the mean radius of the Earth's orbit) is \(1.4960 \cdot 10^{11} \mathrm{~m},\) and a year is \(3.1557 \cdot 10^{7}\) s. Newton's gravitational constant is \(G=\) \(6.6743 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2} .\) Calculate the mass of the Sun in kilograms. (Recalling or looking up the mass of the Sun does not constitute a solution to this problem.)

Short answer

Answer: The approximate mass of the Sun is \(1.989 \cdot 10^{30}\) kilograms.

Step-by-step solution

Write down Kepler's Third Law of Planetary Motion

Kepler's Third Law states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit, with the constant of proportionality being determined by the masses of the two celestial bodies involved and Newton's gravitational constant (G): \(T^2 = \frac{4\pi^2}{G(M_1 + M_2)} \cdot a^3\) Since the mass of the Earth (\(M_2\)) is much smaller than the mass of the Sun (\(M_1\)), for this problem, we can approximate the equation as: \(T^2 \approx \frac{4\pi^2}{GM_1} \cdot a^3\)

Insert the given values in the equation

Now let's insert the given values for the astronomical unit (AU), the length of a year in seconds (T), and Newton's gravitational constant (G) into the equation we derived in Step 1: \((3.1557 \cdot 10^{7} \mathrm{~s})^2 \approx \frac{4\pi^2}{(6.6743 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2})(M_1)} \cdot (1.4960 \cdot 10^{11} \mathrm{~m})^3\)

Solve for the mass of the Sun (M_1)

Now we need to solve the equation for the mass of the Sun (\(M_1\)). We first simplify the equation: \(M_1 \approx \frac{4\pi^2 (1.4960 \cdot 10^{11} \mathrm{~m})^3}{(6.6743 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2})(3.1557 \cdot 10^{7} \mathrm{~s})^2}\) Then we calculate the value of \(M_1\): \(M_1 \approx 1.989 \cdot 10^{30} \mathrm{~kg}\) So the mass of the Sun is approximately \(1.989 \cdot 10^{30}\) kilograms.

Key concepts

Astronomical Unit (AU)

The Astronomical Unit (AU) is a fundamental measurement used in astronomy to describe vast distances within our solar system. Specifically, one AU represents the mean distance from the Earth to the Sun, which is approximately 149.6 million kilometers or precisely, \(1.4960 \times 10^{11}\) meters.

This unit simplifies the description of distances in space, where traditional units like meters and kilometers would be cumbersome due to the sheer scale. The AU makes it easier to compare distances between celestial objects, such as planets and other entities within our solar system.

By using the AU, astronomers can effectively communicate and calculate orbital parameters of planets. In the context of Kepler's Third Law, the AU helps to determine the semi-major axis of Earth's orbit, which is essential for computing the mass of the Sun.

Gravitational Constant (G)

The Gravitational Constant, often denoted as \(G\), is a key component of Newton's Law of Universal Gravitation. It quantifies the strength of the gravitational force between two masses. The value of \(G\) is \(6.6743 \times 10^{-11} \text{ m}^3\text{ kg}^{-1}\text{ s}^{-2}\).

This constant allows us to calculate the gravitational attraction between two masses, a distance apart. It is crucial for understanding and predicting celestial phenomena, such as the orbits of planets, moons, and artificial satellites.

In the exercise, \(G\) is used in Kepler’s Third Law to relate the orbital period and the semi-major axis of a planet orbiting the Sun to derive the mass of the Sun. The constant provides the necessary balance in the equation, enabling accurate cosmic calculations on such grand scales.

Mass of the Sun

The mass of the Sun is an essential astronomical parameter that impacts the orbits of planets in our solar system. Through the application of Kepler's Third Law and the previously discussed parameters—Astronomical Unit and Gravitational Constant—we can derive the mass of the Sun.

In the specified exercise, by simplifying the relationship given by Kepler’s Third Law \(T^2 \approx \frac{4\pi^2}{GM_1} \cdot a^3\), where \(T\) is the orbital period, \(a\) is the semi-major axis and \(M_1\) is the mass of the Sun, we can solve for \(M_1\).

Using the provided values, the resulting mass of the Sun calculated is approximately \(1.989 \times 10^{30}\) kilograms. This immense mass plays a critical role not only in maintaining the planets' orbits but also in the functioning of solar phenomena and energy emission.